$\mathbb{Z}_p$ is a complete metric space in which $\mathbb{Z}$ is dense

Question

I don't quite understand a few parts of the proof of proposition $3$.

• What is meant by "the ideals $p^n\mathbb{Z}_p$ form a basis of neighborhoods of $0$"? After reading the definition of a neighborhood basis, I'm still uncertain what the message is here.
• Why does $\lim.y_n = x$ prove that $\mathbb{Z}$ is in dense $\mathbb{Z}_p$? I ask partly because I'm not familiar what this limit notation is.

Notation.—Let $x$ be a nonzero element of $\Bbb Z_p$; write $x$ in the form $p^nu$ with $u\in\Bbb U$. The integer $n$ is called the p-adic valuation of $x$ and denoted by $v_p(x)$. We put $v_p(0)=+\infty$ and we have $$v_p(xy)=v_p(x)+v_p(y),\quad v_p(x+y)\ge\inf(v_p(x),v_p(y))$$ It follows easily from these formulas that $\Bbb Z_p$ is an integral domain.

Proposition $\mathbf{3}.$—The topology on $\Bbb Z_p$ can be defined by the distance $$d(x,y)=e^{-v_p(x-y)}\;.$$ The ring $\Bbb Z_p$ is a complete metric space in which $\Bbb Z$ is dense.

The ideals $p^n\Bbb Z_p$ form a basis of neighborhoods of $0$; since $x\in p^n\Bbb Z_p$ is equivalent to $v_p(x)\ge n$, the topology on $\Bbb Z_p$ is defined by the distance $d(x,y)=e^{-v_p(x-y)}$. Since $\Bbb Z_p$ is compact, it is complete. Finally, if $x=(x_n)$ is an element of $\Bbb Z_p$, and if $y_n\in\Bbb Z$ is such that $y_n\equiv x_n\pmod{p^n}$, then $\lim.y_n=x$, which proves that $\Bbb Z$ is dense in $\Bbb Z_p$.

(Original image of text here.)

Answer 1

1. It means that if you take any open subset $0\in U \subset\mathbb Z_p$, then $\exists n$ such that $p^n\mathbb Z_p\subset U$, because $p^n\mathbb Z_p=\{x:d(x,0)\le e^{-n}\}$. You can define topology by the metric, but not conversely, i think.

2. I don't know what this dot means. Just check: $d(x, y_n)\le e^{-n}$, so $\lim d(x, y_n)=0$, therefore $\lim y_n=x$.

Answer 2

A basis for a topology is a set $\mathcal B = \{U_i\}$ of open sets such that any open set $U$ can be written as a union of elements of $\mathcal B$. It is a set that "generates the topology". The point is that if we have an easy to understand basis for our topology, then we can prove things about the entire topology just by looking at the basis.

In this case, we want to prove that the topology on $\mathbb Z_p$ can arise from a certain metric. The first thing to spot is that since $\mathbb Z_p$ is a topological ring (i.e. the group actions are continuous), we only need to show this for open sets containing $0$, since we can just translate this result to other points.

To show that this metric defines every open neighbourhood of $0$ directly would be very fiddly. However, we can easily show it for the basis $\{p^n\mathbb Z_p : n\in\mathbb Z_{\ge0}\}$; since all other open neighbourhoods of $0$ are formed by unions of this basis, we're done.

As for the second part, to show that $\mathbb Z$ is dense in $\mathbb Z_p$, we need to show that the closure of $\mathbb Z$ in the $p$-adic topology is $\mathbb Z_p$. One way of doing this is to show that every element $y\in\mathbb Z_p$ is a limit of a sequence in $\mathbb Z$. The limit notation here is just saying that if we write $x=(x_n)\in \mathbb Z_p$ and choose $y_n \in \mathbb Z$ with $$y_n\equiv x_n\pmod {p^n},$$then$$\lim_{n\to\infty}y_n = x$$so that $x$ is indeed a limit of elements of $\mathbb Z$. The dot just seems to be some shorthand for $n\to\infty$.

Answer 3

Note that $p^n \mathbb{Z}_p$ are all open subgroups of $\mathbb{Z}_p$ (hence open subgroups of $\mathbb{Q}_p$, since $\mathbb{Z}_p$ is open in $\mathbb{Q}_p$), and $$\mathbb{Z}_p \supseteq p \mathbb{Z}_p \supseteq p^2\mathbb{Z}_p \supseteq \cdots \{0\}$$When they say "The ideals $p^n \mathbb{Z}$ form a basis of neighborhoods of $0$," what they mean is that given any open set $V$ containing $0$, you can find a large enough $n$ so that $p^n \mathbb{Z} \subseteq V$. It follows that open sets of the form $$x + p^n \mathbb{Z}_p$$ for $x \in \mathbb{Q}_p$ and $n \geq 0$ (translations of open sets are open) form a basis for the topology on $\mathbb{Q}_p$. Given any open set $V$, for each $v \in V$ you can find a large number $n_v$ such that $v + p^{n_v} \mathbb{Z}_p \subseteq V$, and it follows that $$V = \bigcup\limits_{v \in V} v + p^{n_v} \mathbb{Z}_p$$ and actually, on account of the nonarchimedean metric, you can choose all those open sets to be disjoint (do you see why? For $n \geq 0$, $v + p^n \mathbb{Z}_p$ is literally the open ball with center $v$ and radius $\frac{1}{p^n}$, and any two open balls are either disjoint or one contains the other).

If $X$ is a metric space and $A \subseteq X$, the usual definition of "$A$ is dense in $X$" is that $\overline{A} = X$. This is the same as saying that for any $y \in X$, there exists a sequence $x_n \in A$ such that $x_n \to a$.